I am currently trying to learn Patterson-Sullivan theory, but I am getting stuck on basic questions about ergodic theory. Here is one of them, given as an exercise in one of the texts I am trying to read. If you need wider context, the text (in French) is here, with the relevant question towards the bottom of page 17 (next-to-last sentence of the long paragraph).

Let ($\Omega$, $\Gamma$, $\mu$) be a dynamical system satisfying the following assumptions:

  • $\Omega$ is a Hausdorff, locally compact and $\sigma$-compact topological space;
  • $\Gamma$ is a countable group acting on $\Omega$;
  • $\mu$ is a Radon measure on $\Omega$ that is quasi-invariant by $\Gamma$ (for every $g \in \Gamma$, $g_* \mu$ is absolutely continuous with respect to $\mu$).

Let us recall some definitions:

  • a subset $W \subset \Omega$ is called wandering if, for $\mu$-almost all $w \in W$, the intersection of the orbit $\Gamma w$ with $W$ is finite;
  • we say that the system is completely conservative if $\Omega$ has no wandering subsets of positive measure;
  • we say that the system is completely dissipative if $\Omega$ has some wandering subset $W$ such that $\Omega = \bigcup_{g \in \Gamma} g W$.

Statement to prove: If the system is ergodic and the measure $\mu$ does not have any atoms, then the system is completely conservative.

Even though it is supposed to be an easy exercise ("on vérifie aisément que..."), I am at a loss. I have found a partial proof, that works only if you can assume that the set $W$ introduced in the proof is closed. Unfortunately, I have no idea whether it is true, in general, that every completely dissipative dynamical system is generated by a closed wandering set. Can someone give me a proof of this last statement, or find some other way to fix my proof presented below?

In fact, I have remarkably little intuition about what "completely dissipative" actually means. I am tempted to think that a completely dissipative action should in particular be properly discontinuous, but apparently this is false (though I have never seen an actual counterexample). I would be very happy if I could find, somewhere, an arsenal of examples and counterexamples of completely dissipative and completely conservative systems, that is sufficiently well-stocked to forge an intuition about these properties and to be able to test whether some reasonable-sounding statement is actually true.

The partial proof: Let $\Omega = \Omega_C \sqcup \Omega_D$ be the Hopf decomposition of $\Omega$, i.e. a partition of $\Omega$ into two $\Gamma$-invariant subsets such that $\Omega_C$ is completely conservative and $\Omega_D$ is completely dissipative. Since the system is ergodic, they cannot both have positive measure: this means that $\Omega$ is either completely conservative or completely dissipative. By contradiction, assume the latter: let $W \subset \Omega$ be a wandering subset such that $$\Omega = \bigcup_{g \in \Gamma} g W.$$ Note that the set $$\{x \in W \mid \Gamma x \cap W \text{ is finite}\} \cap \operatorname{Supp}(\mu)$$ has the same (positive) measure as $W$, hence is nonempty. Let $x_0$ be any point in this set. Let $F := \{ g \in \Gamma \mid g x_0 \in W \}$; this is a finite set by construction.

Since $x_0 \in \operatorname{Supp}(\mu)$, every neighborhood of $x_0$ has positive measure; on the other hand, since the measure is Radon and without atoms, any point has neighborhoods of arbitrarily small measure. Let $U$ be some neighborhood of $x_0$ such that for every $g \in F$, we have $\mu(g U) < \frac{\mu(W)}{\# F}$. We can also assume (provided that $W$ is closed!) that for all $g \not\in F$, the image $g U$ is disjoint from $W$. Then consider the set $X := \bigcup_{g \in \Gamma} g U$:

  • this set is $\Gamma$-invariant by construction;
  • it has positive measure by construction;
  • its complement $\Omega \setminus X$ contains in particular $W \setminus X$, the complement of $W \cap X$ in $W$. But by construction, the intersection $W \cap X$ is contained in $\bigcup_{g \in F} g U$, so that we have $$\mu(W \cap X) \leq \sum_{g \in F} \mu(g U) < \mu(W).$$ Hence the complement $W \setminus X$, and hence also the complement $\Omega \setminus X$, has positive measure. But these three properties of the set $X$ contradict ergodicity, QED.

2 Answers 2


The desired statement is entirely measure-theoretic, so it's not really necessary to think about the topology of $X$. By standard descriptive set theory, there exists a Borel linear ordering $\preceq$ on $X$. (This means that the graph of $\preceq$ is a Borel subset of $X \times X$.)

Suppose toward a contradiction that $\mu$ is nonatomic and the $\Gamma$-action is ergodic but that there exists a wandering set $W$ of positive measure. Without loss of generality we may assume that the intersection of every $\Gamma$-orbit with $W$ is finite. The ordering $\preceq$ restricts to a linear ordering on each $\Gamma$-orbit. If a $\Gamma$-orbit $\mathcal{O}$ intersects $W$ then $\mathcal{O} \cap W$ is nonempty and finite, so that the $\preceq$-least element of $\mathcal{O} \cap W$ is well-defined. Since $\preceq$ is Borel, the set $V$ consisting of the $\preceq$-least element of each such set $\mathcal{O} \cap W$ is Borel. The set $V$ contains at most one point from each $\Gamma$-orbit. Since $W \subseteq \bigcup_{\gamma \in \Gamma} V$ and $\mu$ is $\Gamma$-quasi-invariant it follows that $\mu(V) > 0$.

Since $\mu$ is nonatomic, there exists a Borel set $V_1 \subseteq V$ such that if we write $V_2 = V \setminus V_1$ then $\min(\mu(V_1),\mu(V_2)) > 0$. Let $\widetilde{V}_j = \bigcup_{\gamma \in \Gamma} \gamma V_j$. For each $j \in \{1,2\}$ the set $\widetilde{V}_j$ is $\Gamma$-invariant and since $\widetilde{V}_j$ contains $V_j$ we have $\mu(\widetilde{V}_j) > 0$. By ergodicity it follows that $\mu(X \setminus \widetilde{V}_j) = 0$ for each $j \in \{1,2\}$.

We claim that $\widetilde{V}_1$ and $\widetilde{V}_2$ are pairwise disjoint, which will lead to the contradiction $$0 < \mu(\widetilde{V}_2) \leq \mu(X \setminus \widetilde{V}_1) =0$$

We will establish the claim by showing that that existence of $x \in \widetilde{V}_1 \cap \widetilde{V}_2$ leads to a further contradiction. Suppose that there exists such a point $x$. Then there exist $\gamma_1,\gamma_2 \in \Gamma$ with $\gamma_1x \in V_1$ and $\gamma_2x \in V_2$. Since $V_1 \cap V_2 = \emptyset$ we have $\gamma_1x \neq \gamma_2x$. Since $\gamma_1x = \gamma_1\gamma_2^{-1}\gamma_2x$ we see that $\gamma_1x$ and $\gamma_2x$ are in the same $\Gamma$-orbit. This cannot be the case since $\gamma_1x$ and $\gamma_2x$ are both elements of $V$ and $V$ contains at most one point from each $\Gamma$-orbit.

$\mathbf{Response\,\, to \,\,comment}$

The way to see that the set $V$ is Borel is to regard Borel sets as being defined by logical formulas. We say that a formula 'is Borel' if it defines a Borel subset of $X$. Countable existential quantifiers correspond to countable unions and not corresponds to complements. Thus these logical operations preserve the property of being a Borel formula.

It is implicit in the hypotheses that $\Gamma$ acts by Borel automorphisms. Since $\preceq$ is Borel, it follows that the predicate $\gamma x \prec x$ is Borel for any $\gamma \in \Gamma$. The set $V$ can be defined as $$\{x \in W: (\not \exists)(\gamma \in \Gamma)((\gamma x \in W) \wedge (\gamma x \prec x))\}$$.

  • $\begingroup$ Nice proof, thank you! One point remains non-obvious for me: "Since $\preceq$ is Borel, the set $V$ consisting of the $\preceq$-least element of each such set $\mathcal{O} \cap W$ is Borel." Could you elaborate on this? $\endgroup$ Commented Apr 26, 2019 at 8:25
  • $\begingroup$ I responded as an edit to the answer. $\endgroup$ Commented Apr 26, 2019 at 8:57

First a general comment. Unfortunately, the categorical approach is completely missing in the way the measure theory is still taught nowadays, and the fact that there actually is only one "reasonable" purely non-atomic probability measure space (so called Lebesgue, Lebesgue--Rokhlin, or standard probability space) isomorphic to the unit interval endowed with the classical Lebesgue measure is not very widely known. Any Borel probability measure on a Polish space makes it a Lebesgue space, so that in particular all measure spaces one has to deal with in the Patterson theory are from this class.

The definition of a wandering set you are using is a non-standard one. According to the usual definition a measurable set is wandering if all its translates are pairwise disjoint mod 0 (i.e., the intersection with almost any orbit consists of at most one point). However, it is very easy to see that the existence of a wandering set $W$ in your sense implies the existence of a "usual" wandering set $W'$. Namely, realize your measure space as the unit interval, and then for each orbit take the minimal point from its intersection with $W$. This operation is clearly measurable and provides you with a measurable set $W'\subset W$, which is wandering in the usual sense.

You are asking why ergodicity implies complete conservativity. The reason (with the right definition of wandering sets) is very simple: the union of the translates of any non-trivial subset of a wandering set is obviously a non-trivial invariant set. Actually, more generally, the union of the free discrete ergodic components of the action is precisely the dissipative part of the Hopf decomposition. See Kaimanovich for more details and Grigorchuk - Kaimanovich - Nagnibeda for various examples concerning the relationship between dissipativity and minimality of boundary actions.


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